Đáp án:
Giải thích các bước giải:
Áp dụng:
\[{\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\]
Ta có:
\[\begin{array}{l}
{\left( {1 + 2014 - \frac{{2014}}{{2015}}} \right)^2} = {1^2} + {2014^2} + {\left( { - \frac{{2014}}{{2015}}} \right)^2} + 2.1.2014 + 2.1.\left( { - \frac{{2014}}{{2014}}} \right) + 2.2014.\left( { - \frac{{2014}}{{2015}}} \right)\\
= {1^2} + {2014^2} + \frac{{{{2014}^2}}}{{{{2015}^2}}} + 2.2014 - 2.\frac{{2014}}{{2015}}\left( {1 + 2014} \right)\\
= {1^2} + {2014^2} + \frac{{{{2014}^2}}}{{{{2015}^2}}} + 2.2014 - 2.2014\\
= {1^2} + {2014^2} + \frac{{{{2014}^2}}}{{{{2015}^2}}}\\
\Rightarrow \sqrt {{1^2} + {{2014}^2} + \frac{{{{2014}^2}}}{{{{2015}^2}}}} + \frac{{2014}}{{2015}} = 1 + 2014 - \frac{{2014}}{{2015}} + \frac{{2014}}{{2015}}\\
= 2015
\end{array}\]
